1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+0+0+0+3+4+1+2+3+111+44+99 = 357

1+2+3+4+5+6+7+8+9+1+2+3+4+5+6+7+8+9+0+0+0+3+4+1+2+3+111+44+99=357

1+2+3+4+5+6+7+8+9+11+22+33+44+55+66+77+88+99=540

1 + 2 + 3 + 4 + 5 + 6 + 7 + 8+ 9 + 11 + 22 + 33 + 44 + 55 + 66 + 77 + 88+ 99

= ( 1 + 2 + 3 + 4+ 5 + 6 + 7 + 8 + 9 ) + ( 11 + 22 + 33 + 44 + 55 +66 +77 + 88 + 99)

= 45                                                 +                   495

=      540

a) \(A=\left\{1;2;3;4;5\right\}\)

\(\Rightarrow A=\left\{x\inℕ|1\le x\le5\right\}\)

b) \(B=\left\{0;1;2;3;4\right\}\)

\(\Rightarrow B=\left\{x\inℕ|0\le x\le4\right\}\)

c) \(C=\left\{1;2;3;4\right\}\)

\(\Rightarrow C=\left\{x\inℕ|1\le x\le4\right\}\)

d) \(D=\left\{0;2;4;6;8\right\}\)

\(\Rightarrow D=\left\{x\inℕ|x=2k;0\le k\le4;k\inℕ\right\}\)

e) \(E=\left\{1;3;5;7;9;...49\right\}\)

\(\Rightarrow E=\left\{x\inℕ|x=2k+1;0\le k\le24;k\inℕ\right\}\)

f) \(F=\left\{11;22;33;44;...99\right\}\)

\(\Rightarrow F=\left\{x\inℕ|x=11k;1\le k\le9;k\inℕ\right\}\)

Bngxgyfiyfyg

Tui không biết nha giúp tui làm

550 bạn nhé

hoi vo van

đúng òi vớ va vớ vẩn

\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\frac{22}{44}\)

\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\frac{22}{44}\)

\(=\frac{3}{5}+\frac{3}{7}-\frac{1}{35}+\frac{3}{11}-\frac{3}{4}+\frac{22}{44}+\frac{2}{97}\)

\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}+\frac{22}{44}\right)+\frac{2}{97}\)

\(=\left(\frac{21}{35}+\frac{15}{35}-\frac{1}{35}\right)+\left(\frac{12}{44}-\frac{33}{44}+\frac{22}{44}\right)+\frac{2}{97}\)

\(=\frac{35}{35}+\frac{1}{44}+\frac{2}{97}\)

\(=1+\left(\frac{1}{44}+\frac{2}{97}\right)\)

\(=1+\left(\frac{97}{4268}+\frac{88}{4268}\right)\)

\(=1+\frac{185}{4268}=1\frac{185}{4268}\)

Chúc bạn học tốt

Trả lời:

\(\frac{3}{5}+\frac{3}{11}-\left(\frac{-3}{7}\right)+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\frac{22}{44}\)

\(=\frac{3}{5}+\frac{3}{11}+\frac{3}{7}+\frac{2}{97}-\frac{1}{35}-\frac{3}{4}+\frac{22}{44}\)

\(=\left(\frac{3}{5}+\frac{3}{7}-\frac{1}{35}\right)+\left(\frac{3}{11}-\frac{3}{4}+\frac{22}{44}\right)+\frac{2}{97}\)

\(=1+\frac{1}{44}+\frac{2}{97}\)

\(=\frac{4453}{4268}\)

`@` `\text {Ans}`

`\downarrow`

`e)`

`1+(-4)+7+(-10)+13`

`= (-4 - 10) + (1 + 13) + 7`

`= -14+14 + 7`

`= 7`

`f)`

`-2+7+(-12)+17+(-22)`

`= (7 + 17) + (-12 - 22) - 2`

`= 24 - 34 - 2`

`= -10 - 2`

`= -12`

`g)`

`44+45+46+47-24-25-26-27`

`= (44 - 24) + (45 - 25) + (46 - 26) + (47 - 27)`

`= 20 + 20 + 20 + 20`

`= 20*4 = 80`

a) \(\dfrac{2}{3}x-\dfrac{1}{2}=\dfrac{1}{10}\)

\(\dfrac{2}{3}x=\dfrac{1}{10}+\dfrac{1}{2}=\dfrac{3}{5}\)

\(x=\dfrac{3}{5}:\dfrac{2}{3}=\dfrac{9}{10}\)

b) \(\dfrac{39}{7}:x=13\)

\(x=\dfrac{\dfrac{39}{7}}{13}=\dfrac{3}{7}\)

c) \(\left(\dfrac{14}{5}x-50\right):\dfrac{2}{3}=51\)

\(\dfrac{14}{5}x-50=51\cdot\dfrac{2}{3}=34\)

\(\dfrac{14}{5}x=34+50=84\)

\(x=\dfrac{84}{\dfrac{14}{5}}=30\)

d) \(\left(x+\dfrac{1}{2}\right)\left(\dfrac{2}{3}-2x\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x+\dfrac{1}{2}=0\\\dfrac{2}{3}-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{2}\\x=\dfrac{1}{3}\end{matrix}\right.\)

e) \(\dfrac{2}{3}x-\dfrac{1}{2}x=\dfrac{5}{12}\)

\(\dfrac{1}{6}x=\dfrac{5}{12}\)

\(x=\dfrac{5}{12}:\dfrac{1}{6}=\dfrac{5}{2}\)

g) \(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\dfrac{11}{5}-\dfrac{3}{7}=-2\)

\(\left(x\cdot\dfrac{44}{7}+\dfrac{3}{7}\right)\cdot\dfrac{11}{5}=-2+\dfrac{3}{7}=-\dfrac{11}{7}\)

\(x\cdot\dfrac{44}{7}+\dfrac{3}{7}=-\dfrac{11}{7}:\dfrac{11}{5}=-\dfrac{5}{7}\)

\(\dfrac{44}{7}x=-\dfrac{5}{7}-\dfrac{3}{7}=-\dfrac{8}{7}\)

\(x=-\dfrac{8}{7}:\dfrac{44}{7}=-\dfrac{2}{11}\)

h) \(\dfrac{13}{4}x+\left(-\dfrac{7}{6}\right)x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x-\dfrac{5}{3}=\dfrac{5}{12}\)

\(\dfrac{25}{12}x=\dfrac{5}{12}+\dfrac{5}{3}=\dfrac{25}{12}\)

\(x=1\)

Mỏi tay woa bn làm nốt nha!!

cau 1:b,cau2d,cau3b

a) \(A=111...1555...56\) (n cs 1, n-1 cs 5)

\(A=111...1000...0+555...50+6\) (n cs 1, n cs 0 (không tính số 0 ở số 555...50), n-1 cs 5)

\(A=111...1.10^n+555...5.10+6\) (n cs 1, n-1 cs 5)

\(A=\dfrac{999...9}{9}.10^n+\dfrac{5}{9}.999...9.10+6\) (n cs 9 ở phân số thứ nhất, n-1 cs 9 ở phân số thứ 2)

\(A=\dfrac{10^n-1}{9}.10^n+\dfrac{5}{9}.\left(10^{n-1}-1\right).10+6\)

\(A=\dfrac{\left(10^n\right)^2-10^n+5.10^n-50+54}{9}\)

\(A=\dfrac{\left(10^n\right)^2+4.10^n+4}{9}\)

\(A=\left(\dfrac{10^n+2}{3}\right)^2\)

 Hiển nhiên \(3|10^n+2\) vì \(10^n+2\) có tổng các chữ số bằng 3, suy ra A là số chính phương.

Câu b áp dụng kĩ thuật tương tự nhé bạn.